Consider the reaction: N2 + 3H2 → 2NH3 carried out at constant temperature and pressure. If Δ H and Δ U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

Question

Consider the reaction: N2 + 3H2 → 2NH3 carried out at constant temperature and pressure. If Δ H and Δ U are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ? (A)  Δ H > Δ U (B)  Δ H < Δ U (C)  Δ H = Δ U (D)  Δ H = 0

Tardigrade Admin · Accepted Answer

Δ H = Δ U + Δ ngRT Δ H = enthalpy change (at constant pressure) Δ U = internal energy change (at constant volume) (given reaction is exothermic) Thus Δ H < Δ U. Δ ng = mole of (gaseous products - gaseous reactants) = - ve Thus Δ H < Δ U. Note: Numerical value of Δ H < Δ U in exothermic reaction and when Δ ng < 0

Q. Consider the reaction : $N_{2} + 3 H_{2} \to 2 N H_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ U$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

$Δ H > Δ U$

$Δ H < Δ U$

$Δ H = Δ U$

$Δ H = 0$

Q. Consider the reaction : N2​+3H2​→2NH3​ carried out at constant temperature and pressure. If ΔH and ΔU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

ΔH>ΔU

ΔH<ΔU

ΔH=ΔU

ΔH=0

Q. Consider the reaction : $N_{2} + 3 H_{2} \to 2 N H_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ U$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true ?

$Δ H > Δ U$

$Δ H < Δ U$

$Δ H = Δ U$

$Δ H = 0$